Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given a non-empty set $S$ and a Banach space $X$. Let $B(S,X)$ be the space of all bounded maps from $S$ to $X$. Can we identify $B(S,X)$ with $\ell^\infty(S) \otimes X$, where $\otimes$ is some kind of tensor product of Banach spaces?

share|cite|improve this question

I don't think so.

A tensor product of Banach spaces generally means a completion of the algebraic tensor product under some norm. In this case the algebraic tensor product $\ell^\infty(S) \otimes_a X$ embeds naturally into $B(S,X)$, whose norm is given. So the question becomes, whether $\ell^\infty(S) \otimes_a X$ is dense in $B(S,X)$, and the answer in general is no.

Let $S = \mathbb{N}$ and let $X$ be an infinite dimensional separable Banach space. Choose a sequence $x_n$ which is dense in the unit sphere of $X$. Define $f : S \to X$ by $f(n) = x_n$, so $f \in B(S,X)$. But for any $g \in \ell^\infty(S) \otimes_a X$, the range of $g$ is contained in a finite dimensional subspace $E_g$ of $X$. By Hahn-Banach we can find $x$ in the unit sphere with $d(x, E_g) > 1/2$, and by density there is $x_n$ with $||x_n - x|| < 1/4$. Thus $\sup_n ||g(n) - f(n)|| > 1/4$ so $f$ is not in the closure of $\ell^\infty(S) \otimes X$.

share|cite|improve this answer
Brilliant! Thank you. – balzac Jun 21 '11 at 18:10
Thus, there is no better description of $B(S,X)$ than $C(\beta S, X)$, where $S$ carries the discrete topology... – balzac Jun 21 '11 at 18:12
@balzac: This might be a very naive question, but why does $C(\beta S, X) = B(S,X)$? The universal property of $\beta S$ guarantees that continuous functions on $S$ valued in a compact space extend to $\beta S$, but bounded subsets of $X$ need not be precompact. – Nate Eldredge Jun 21 '11 at 18:53
You're right: I am wrong. Btw, correct me if I am wrong: if $S$ is uncountable then there exist $f\in B(S,\mathbb{R})$ which is not a uniform limit of functions taking finitely many values? – balzac Jun 21 '11 at 19:19
@balzac: No. Any $f \in B(S, \mathbb{R})$ is a uniform limit of functions taking finitely many values. For instance, say $0 \le f \le 1$; then if $f(x) \in [k/n, (k+1)/n)$, set $f_n(x) = k/n$, and we have $||f - f_n||_\infty \le 1/n$. More generally, this works for any function taking values in a totally bounded metric space. – Nate Eldredge Jun 21 '11 at 19:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.